MIMO with Zero Forcing equalizer

by Krishna Sankar on October 24, 2008

We had discussed three Single Input Multiple Output (SIMO also known as receive diversity) schemes – Selection combining, Equal Gain Combining, Maximal Ratio Combining and a Multiple Input Single Output (MISO, also known as transmit diversity) scheme – Alamouti 2×1 STBC. Let us now discuss the case where there a multiple transmit antennas and multiple receive antennas resulting in the formation of a Multiple Input Multiple Output (MIMO) channel. In this post, we will restrict our discussion to a 2 transmit 2 receive antenna case (resulting in a 2×2 MIMO channel). We will assume that the channel is a flat fading Rayleigh multipath channel and the modulation is BPSK.

2×2 MIMO channel

In a 2×2 MIMO channel, probable usage of the available 2 transmit antennas can be as follows:

1. Consider that we have a transmission sequence, for example

2. In normal transmission, we will be sending in the first time slot, in the second time slot, and so on.

3. However, as we now have 2 transmit antennas, we may group the symbols into groups of two. In the first time slot, send and from the first and second antenna. In second time slot, send and from the first and second antenna, send and in the third time slot and so on.

4. Notice that as we are grouping two symbols and sending them in one time slot, we need only time slots to complete the transmission – data rate is doubled !

5. This forms the simple explanation of a probable MIMO transmission scheme with 2 transmit antennas and 2 receive antennas.

Having said this, some of you will wonder – the two transmitted symbols interfered with each other. Can we ever separate the two out? The rest of the post attempts to answer this question.

Figure: 2 Transmit 2 Receive (2×2) MIMO channel

Other Assumptions

1. The channel is flat fading – In simple terms, it means that the multipath channel has only one tap. So, the convolution operation reduces to a simple multiplication. For a more rigorous discussion on flat fading and frequency selective fading, may I urge you to review Chapter 15.3 Signal Time-Spreading from [DIGITAL COMMUNICATIONS: SKLAR]

2. The channel experience by each transmit antenna is independent from the channel experienced by other transmit antennas.

3. For the transmit antenna to receive antenna, each transmitted symbol gets multiplied by a randomly varying complex number . As the channel under consideration is a Rayleigh channel, the real and imaginary parts of are Gaussian distributed having mean and variance .

4. The channel experienced between each transmit to the receive antenna is independent and randomly varying in time.

5. On the receive antenna, the noise has the Gaussian probability density function with

with and .

7. The channel is known at the receiver.

Zero forcing (ZF) equalizer for 2×2 MIMO channel

Let us now try to understand the math for extracting the two symbols which interfered with each other. In the first time slot, the received signal on the first receive antenna is,

.

The received signal on the second receive antenna is,

.

where

, are the received symbol on the first and second antenna respectively,

is the channel from transmit antenna to receive antenna,

is the channel from transmit antenna to receive antenna,

is the channel from transmit antenna to receive antenna,

is the channel from transmit antenna to receive antenna,

, are the transmitted symbols and

is the noise on receive antennas.

We assume that the receiver knows , , and . The receiver also knows and . The unknown s are and . Two equations and two unknowns. Can we solve it? Answer is YES.

For convenience, the above equation can be represented in matrix notation as follows:

.

Equivalently,

To solve for , we know that we need to find a matrix which satisfies . The Zero Forcing (ZF) linear detector for meeting this constraint is given by,

.

This matrix is also known as the pseudo inverse for a general m x n matrix.

The term,

.

BER with ZF equalizer with 2×2 MIMO

Note that the off diagonal terms in the matrix are not zero (Recall: The off diagonal terms where zero in Alamouti 2×1 STBC case). Because the off diagonal terms are not zero, the zero forcing equalizer tries to null out the interfering terms when performing the equalization, i.e when solving for the interference from is tried to be nulled and vice versa. While doing so, there can be amplification of noise. Hence Zero Forcing equalizer is not the best possible equalizer to do the job. However, it is simple and reasonably easy to implement.

Further, it can be seen that, following zero forcing equalization, the channel for symbol transmitted from each spatial dimension (space is antenna) is a like a 1×1 Rayleigh fading channel (Refer Section 3.3 of [WIRELESS-TSE, VISWANATH]). Hence the BER for 2×2 MIMO channel in Rayleigh fading with Zero Forcing equalization is same as the BER derived for a 1×1 channel in Rayleigh fading.

Summary

1. As expected, the simulated results with a 2×2 MIMO system using BPSK modulation in Rayleigh channel is showing matching results as obtained in for a 1×1 system for BPSK modulation in Rayleigh channel.

2. As noted in Section 3.3 of [WIRELESS-TSE, VISWANATH], the Zero Forcing equalizer is not the best possible way to equalize the received symbol. The zero forcing equalizer helps us to achieve the data rate gain, but NOT take advantage of diversity gain (as we have two receive antennas).

3. We might not be able to achieve the two fold data rate improvement in all channel conditions. It can so happen that channels are correlated (the coefficients are almost the same). Hence we might not be able to solve for the two unknown transmitted symbols even if we have two received symbols.

4. It is claimed that there can be receiver structures which enables us to have both diversity gain and data rate gain. In future posts, the attempt will be to discuss receiver structures which hopefully enables us to find out approaches which will help us to keep the data rate gain, but still move from the 1×1 curve to 1×2 MRC curve.

First of all I want to say thank you very much for writing this notes.

I have got a question about adding more antennas at the transmitting and receiving side, eg. 64*64 MIMO systems. I have tried to made modifications based on your programme but I don’t have a clue. Could you please give me some tips on possible modifications? Thank you so much!

First of all thanks a lot for your information on matlab coding. I learnt matlab coding with the help of your codes. Thanks a lot. My question is ” I’m doing project on orthogonal space-time block codes under rayleigh channel. There is a code for this,but for a codeword of 4×4 matrix(rate=1) but i need it for8x4 matrix(rate of 0.5) and other rates of 0.75. I actually tried with QPSK modulation and demodulation, my BER vs SNR performance is of zigzag manner. Then i used your qam16 mod and demod and got 0.01 ber for an snr of 20(I hope this is not good performance)”

@sandeep: well, to rule out any obvious issue, first make sure that the ber is zero when there is no noise added. once that is done, the exact performance number will depend on your coding type which is employed. try to compare with the theoretically estimated ber curves.good luck.

Sir, am a research scholar from south India. i found your blog about 2*2 MIMO & zero forcing equalizer and is very interesting. am quite thankful to u if u suggest me some ideas to modify the code for various ocnfigurations 4*4 ,1*4 .. etc.

please Mr. Krishna Sankar Iwant to build simple mimo ofdm system to simulate snr vs. ber but I want it the mimo system to be flexable not only 2×2 so ican change the number of transmitter and recievers please help me if you have the matlab code please send it to me

@eng_dina: In most of the articles which I have discussed I have used 2×2 MIMO case. And to increase the speed of the Matlab simulations, I have not used inv() operation in Matlab. You can try using inv() operation and increase the speed of the simulations

Thanks for your great and helpful blog, I am wondering if i can extend your code to the case of multiuser MIMO, for two users rather than one? just want to know weather you think it is possible so i can start working on it. big thanks again

I have a question about plotting BER vs SNR in MIMO systems. Lets say i have a 2×2 MIMO system, and i perform an SVD on the channel, so i can encode and decode the signal with V and U, respectively, such that i am left with 2 uncoupled SISO channels. now each channel has its own singular value, and let us assume for now that the noise power is the same in both channels. So we have SNR1 in the first channel, and SNR2 in the second channel, and they are different.
Q1: How would you define the SNR of this system? is it the average SNR of both channels?

Now we now that according to the Water Filling algorithm, we can transmit more power (and thus, more bits) in channels with higher SNR. so if come to conclusion that the first channel can carry QPSK, and the second channel can only carry BPSK, i have different contelations in both channels… So the SNR which were calculated earlier for each channel, should now be modified in order to reflect the number of bits which are transmitted via those channels, by: SNR_bit=SNR_sym/k, where k=log_2(M).

Q2: How should plot 1 graph which describes the BER vs. SNR_bit of this system? Should i calculate SNR_bit for each SISO channel, and BER for each SISO channels separately, and the average the results?

@ben: My replies
1.a) As I recall, when we are multiplying by V, we are not introducing any gain in the transmitter. In that case, signal power at the transmitter remains the same irrespective of multiplication by V. Agree?
Having said that reporting SNR information from both the streams and average SNR is useful. I believe most MIMO test instruments report both
1.b) The signal to noise ratio is independent of the constellation used. If you want to compare using Eb/N0, then yes, need to factor in the constellation
2) Hmm…. you can defined Es/N0 = ratio of total power at tx / total noise power at rx. Then depending on the number of bits which are pumpre through, Es/N0_dB = Eb/N0_dB + 10*log10(number of bits per symbol across all tx chains)

1.a I agree. I want to have a single number which describes the SNR of the system. In that case, i will take the average. The only question left is which average? mean, or geometrical average?In page 12 of http://www.stanford.edu/group/cioffi/publications.html, Cioffi takes the geometric one.

1.b. Since in order to compare two transmission schemes, which uses different constellations, i need to compare the BER (SER is almost meaningless, e.g., if scheme 1 uses 64QAM and scheme 2 uses BPSK….), so the x axis should read SNR_bit (=Eb/No) rather than SNR_sym (=Es/N0).

2. Inside the log is the average bits per symbol over all Tx chaise? so if i Tx QPSK in stream 1, and BPSK in stream 2, than it will be qual to 1.5?
SNR_bit=SNR_sym-10log10(1.5)?

How can I make sure that E[||H|| * ||H||] =1 ?
ie Expectation of square of Mod of the Channel matrix is equal to one.
This is the first condition that has to be met before I start to simulate.
Is the term [||H|| * ||H||] relating to the a single channel realization or an ensemble of them. In anycase how do I generate H satisfying the above condition and verify this the same. H is a Rayleigh fading channel initially and later on (not related to the first one) I will have to look into the cascaded rayleigh fading channel.

lets say I have 2 streams of Data which are coupled using a Q matrix and are transmitted. Then,
Y=H[]*Q[]*S[]
by solving for S[] I am able to calculate the stream EVM
-but how do I calculate channel EVM ?
-is the Q[] matrix known at the receiver
-is it possible to extract Q[] matrix at the receiver

my question is: If we use the same technique (e.g spatial multiplexing with the ZF receiver, or Alamouti scheme ) in the conventional MIMO performing over the flat fading channel and in MIMO-OFDM performing over the frequency selective channel, should be the results same?

I mean,… MIMO is designed for flat fading channels and OFDM transform frequency selective channel into several flat fading channel so the performance should be same, but I am not sure…

@jhon: 1/H is same as H^H/(H^H*H) i.e multiplying by H^H in numerator and denominator. This makes the matrix general for any non square matrix.
Just a quick matlab code snippet:
h = rand(2,2)+j*rand(2,2)
w1 = inv(h’*h)*h’
w2 = inv(h)

hi Krishna Sankar, you do all the simulations in flat-fading environment, how about frequency selective channels which are more realistic than the former? In fact, I use ITU channel models in my simulation but it’s difficult to me to apply your codes with more taps. Would you please help me with that problem. Thank you in advance

Well, i think there is no formula for uncoded BER at multi-tap Rayleigh fading channels
what will change is that if you perform equalization in time domain then you would have to do filtering (convolution) which is of course more complex than simple multiplication for one-tap fading, but if you use systems like LTE, WiMAX (OFDM, SC-FDMA based) where you have cyclic prefix and freq.domain pilots for channel estimation then there is no problem for simple frequency domain equalization.
Another thing, ITU channel were originally for SISO systems, they have no information about spatial correlation between antennas (antenna spacing, polarization, angular spread, AoA etc.). But this model can be extended to MIMO channel models with the definition of a per-tap spatial correlation.

Tow questions about your code:
Q1, I think the channel need to be normalized which means h*conj(h’)=1, but it seems you did not do that, how do you think about the channel matrix normalization as I described.

Q2, To calculate the inverse matrix, in your code, why do not you directly use the matlab ‘inv()’ method but implement your self with a block of code?

@chen: My replies:
1/ The E{h*h’} over all channel realization (and not per realization) is unity.

2/ As I recall, the inv() operator works only a for two dimensional matrix, and I would have to put a for loop to perform the equalization for all symbols. To speeden things up, I used my own inv() and performed the whole story using matrix algebra.

That what was bothering me for some long time – difference between ZF and MRC. So, you mean there is no MRC for MIMO?

@Street hawk: well, if you have correlated fading then probably ML-based receivers would be a good choice. Or you could use correlation in order to perform beamforming to suppress inter-cell interference (like IRC receiver based on e.g. MMSE or MLD).

Hi Mr krishna
I have an question about pseudo inverse matrix;
We know that we want to get an identity matrix from the equation WH=I.
So we can get the same result by using another equation which is
W=inv(H) , because [ inv(H)*H=I ]
(I consider this equation[W= inv(H)*H ] in my work and it gave me the same result.)
Can u pls explain why we can not use such equation??

@sam2: The inv(H) is not defined for a non-square matrix. Hence the pseudo inverse is a more general definition which works for non-square matrix, and scales down to inverse operation for a square matrix.
Agree?

Hi krishna…
Can you please tell me Why you are using
y = squeeze(sum(h.*sMod,2)) + 10^(-Eb_N0_dB(ii)/20)*n;
instead of :-
y = squeeze(sum(h.*sMod,1)) + 10^(-Eb_N0_dB(ii)/20)*n;
becauase the using of sum(s,2) will sum the raws of information bits
like this:-
ip =[0 0 1 1];
s=[-1 -1 1 1];
x=sum(s,2)
will result in x=0 and that will not make any sense to me?

@mohanad: The way the matrix multiplication is organized for each time slot is as follows:
h = [h11 h12 ; h21; h22] and s = [s1 s2;s1 s2]
We know that y = [h11*s1 + h12s2; h21*s1 + h22*s2]
So, how do we achieve that?
We do a dot product of h and s, and the sum in the column dimension. To the sum in the column dimension, we have the parameter sum(,2).

Thanks very much for the reply. The case that am considering is that
y1=x1+x2+n1
y2=x1+x2+n2
the channel is AWGN channel and I want to apply the ZF or MMSE equilizer to de-multiplex the channels. I this case, I assume that h11, h12, h21 and h22 are equal to 1. Therefore, the inverse of the channel matrix will be INF?
If you could help me to understand how to apply the MIMO detectors for AWGN channel?

@SARA: If you have
y1=x1+x2+n1
y2=x1+x2+n2
the you wont be able to solve for x1 and x2, given that you know y1 and y2. Think of your linear algebra classes. To solve for two unknowns (x1, x2), we need two equations (y1, y2). If the two equations carry ‘same/similar’ information, then there is no way that we can use them for resolving y1, y2. In MIMO communication parlance, having such a channel is called rank deficient channel.

The problem you have is that you assume your both MIMO users (if it spatial MIMO case for example) have identical channel. That means both MIMO users are completely correlated. Your channel matrix is [1 1 ; 1 1] => you cant perform pseudoinverse inv(H^H*H). what you could do is set a phase difference between Rx antennas for one user, let’s say its ‘pi’, so your channel would be H=[1 1; 1 -1] => fully uncorrelated. It is a common problem for MMSE/ZF receivers when channels are correlated. Whereas, MLD could still exploit symbol constellation and outperform MMSE/ZF.

@aydar: In general, I agree with the intent of the comment. However, just to add that I have not assumed that the channel matrix is identical. I assume that the channel matrix takes tap from Rayleigh distributed channel.

Hi Krishna,
Thanks very much for the intersting topics. Could I ask what will be the case if we have AWGN channel. In this case h11, h12, h21 and h22 will be 1. and the inverse of H will Inf.
Could you explain in more details how this could be solved for AWGN channel
Regards

@Khattak: Thanks. Well, I have not written posts on the 4×3 or 4×4 MIMO systems. Anyhow, I hope that you will be able to adapt the existing code to the above configurations. Good luck for you algorithm explorations.

Dear Mr. Krishna, you are really helping students a lot. Please, I would like to know, why you used the operation sum(h(:,2,:).*conj(h(:,2,:)),1);to calculate each of the cofficents, because c*conj(c) will result in (abs(C)).^2 or in other worlds the magintude squared of each of the column elements. also, why taking summation?. Thank you again and best wishes.

@communications_engineer: The term H^H is Hermitian operator – in simple terms its the conjugate transpose of a matrix.
If there is only one path, then the equation reduces to H*/|H|^2, where H* is the conjuagate of H. Agree?

Krishna, as I have written before I’m working on CDMA MUD, so I want to use ZF equalization (I now its not good, but only for the sake of it), so I’ll use ZF before the de-correlation (I’m using eq y1 = A1*r2 + A2*r2*sgma+noise, verdu’s book), so ZF before decision making.

@John: Note that we receive two copies of the transmit symbol at the receiver (one on each receive antenna). The presence of two copies helps to achieve the ZF performance eventhough there is interference.
Do you agree?

@John: Yes, there is noise enhancement. But I would think that is case for 1×1 too. Hence 1×1 ZF can compare with 2×2 ZF.

As an additional thought:
If I see 1×1 with ZF equalization and MMSE equalization, I find that the equalized constellaion has a lower EVM/MER in the presence of MMSE. However, the raw BER is the same. I fail to understand that behaviour. Can you please share comments on that aspect?

@ Sky Stradlin: Yes, Zero Forcing and MMSE classifies two different types of receiver structure (with MMSE performing better than Zero Forcing in MIMO case). Can you please point me to the literature which discuss ZF/MMSE receiver.

@karl: Well, recall that in the MIMO case we require only half the time to transmit the symbols.

Assuming P is the transmit power from a single antenna in a 1×1 case and we require time T to send N symbols. The energy
consumed in E_{1×1| = PT
For the 2×2 MIMO, the time taken is only T/2 and we assume that we transmit from both the antennas at power P each. The energy consumed is E_{2×2} = 2P*T/2 = PT
Since the total energy consumed is the same, I think it is fair to compare with the SISO 1×1 case.
Do you agree to my perspective? Kindly share your thoughts

@Jarrod: As is the case with MIMO, if there is ISI, then Zero Forcing equaliser does not be the optimal way to equalize. Conceptually ISI can be thought of as interference in the time domain where as MIMO can be thought of as interference in the spatial domain.

thanks for this blog, I am researching signal processing for MIMO. I have utilized vblast, qostbc, with mmse but I don’t know that algorithm can I utilized for wideband. Could you help me? Do you have some example code for ISI?

A long time ago I used to work in DSP and communications. Since then this world has completely changed. I am looking forward to catching up by reading your blog.

You seem to have put a lot of effort, even have equations in here.

I have started out with this article so I dont know if I have missed some things…but you could make it more easier and interesting for readers like me through real life examples and putting more of your own insights into the equations.

Hi sir,
I got the point for the first two lines, which is ..
yMod = kron(y,ones(1,2)); % formatting the received symbol for equalization
yMod = sum(hMod.*yMod,1); % H^H * y

here the column wise addition is there because of the transpose of the conj(H) matrix with yMod.

But for the second two lines which are..
yMod = kron(reshape(yMod,2,N/nTx),ones(1,2)); % formatting
yHat = sum(reshape(hInv,2,N).*yMod,1); % inv(H^H*H)*H^H*y

Here there is no transpose operation. i.e we are just multiplying inv(H^H*H) and H^H*y. There is no transpose of inv(H^H*H). So instead of those two lines, I think it will be..
hInv = reshape(hInv,[nRx,nTx,N/nTx]);
yMod = kron(yMod,ones(nRx,1)); %
yMod = reshape(yMod,[nRx,nTx,N/nTx]);
yHat = sum(hInv.*yMod,2);
yHat = reshape(yHat,1,N);
I have applied those codes and found better results. Sir I am in dilemma whether I am right or absolutely wrong..Please help me out as soon as possible..
Waiting for your valuable feedback.
Thank You.